Posted on July 31, 2026
A long vertical cylinder has three narrow open rings, each of which wraps around seven-eighths of the cylinder (leaving a one-eighth "gap"). The rings are evenly spaced vertically, but are otherwise randomly rotated about the cylinder's central axis.
Fiddler: For some orientations of the rings, there exists at least one vertical line down the cylinder's surface that passes through each ring's gap. What is the probability that at least one such vertical line exists?
Extra Credit: Instead of requiring a vertical line down the cylinder's surface, now any helix down the surface is allowed. What is the probability that there exists at least one such helix that can pass through each ring's gap?
Let the circumference of the cylinder be \(1\). Let each gap have a length \(f\) (in the problem, \(f = 1/8\)). Let \(y_0, y_1, y_2 \in [0, 1)\) be the center positions of the three gaps, chosen independently and uniformly at random.
A vertical line at position \(x\) passes through all three gaps if and only if there exist offsets \(\epsilon_i \in [-f/2, f/2]\) such that for each ring \(i\):
\[ y_i + \epsilon_i \equiv x \pmod 1 \]
This requirement can be expressed as a system of two independent equations by eliminating \(x\):
\[ \begin{aligned} y_0 - y_1 &\equiv \epsilon_1 - \epsilon_0 \pmod 1 \\ y_1 - y_2 &\equiv \epsilon_2 - \epsilon_1 \pmod 1 \end{aligned} \]
Let \(D_1 = y_0 - y_1\) and \(D_2 = y_1 - y_2\). Since the \(y_i\) are independent and uniform, the pair \((D_1, D_2)\) is uniformly distributed on the unit square (the torus \([0, 1) \times [0, 1)\)). The probability of a vertical line existing is the area of the region in the \((D_1, D_2)\) plane reachable by the error terms \(\Delta_1 = \epsilon_1 - \epsilon_0\) and \(\Delta_2 = \epsilon_2 - \epsilon_1\).
Given \(|\epsilon_i| \le f/2\), the possible values for \((\Delta_1, \Delta_2)\) form a hexagon defined by the constraints:
The area of this hexagon is \(3f^2\). For \(f = 1/8\):
\[ P = 3 \times \left(\frac{1}{8}\right)^2 = \frac{3}{64} = 0.046875. \]
Answer: \(\displaystyle\boxed{\frac{3}{64}}\), or \(0.046875\).
For the extra credit, any helix is allowed. A helix is defined by a linear path \(x(z) = x_{start} + \alpha z \pmod 1\). The helix passes through all three gaps if there exist \(x_{start}\) and \(\alpha\) such that:
\[ y_i + \epsilon_i \equiv x_{start} + i\alpha \pmod 1 \]
for \(i = 0, 1, 2\). By eliminating the two degrees of freedom (\(x_{start}\) and \(\alpha\)), we are left with a single linear constraint on the centers and offsets:
\[ y_0 - 2y_1 + y_2 \equiv 2\epsilon_1 - \epsilon_0 - \epsilon_2 \pmod 1 \]
Let \(D = y_0 - 2y_1 + y_2 \pmod 1\). Since the coefficients \((1, -2, 1)\) are integers with a GCD of \(1\), \(D\) is uniformly distributed on the circle \([0, 1)\). A helix exists if \(D\) falls within the range of the error term \(E = 2\epsilon_1 - \epsilon_0 - \epsilon_2\). Given \(\epsilon_i \in [-f/2, f/2]\), the range of \(E\) is \([-2f, 2f]\).
The length of this interval (and thus the probability) is \(4f\). For \(f = 1/8\):
\[ P = 4 \times \frac{1}{8} = \frac{1}{2}. \]
Note the relationship between the two solutions. In the 3-ring case:
In general, adding flexibility to the path reduces the "effective dimensionality" of the random collision required, significantly increasing the probability of success.
Extra credit answer: \(\displaystyle\boxed{\frac{1}{2}}\).
Both results generalize cleanly to \(n\) rings, each with gap length \(f\) and independently, uniformly random center \(y_0, \ldots, y_{n-1}\).
Vertical line. A vertical line exists if and only if all \(n\) gap-centers fit within some single arc of length \(f\) on the circle. Partition on which point is the "leading" (counterclockwise-most) point of that arc: for each of the \(n\) points, the probability that the other \(n-1\) points all fall within \(f\) of it (going clockwise) is \(f^{n-1}\), and these \(n\) events are disjoint. So:
\[ P_{\text{line}}(n) = n f^{n-1}, \qquad \text{valid for } f \le \tfrac{1}{2}. \]
At \(n = 3\) this is \(3f^2\), matching the Main Solution above.
Helix. Eliminating the helix's two free parameters (\(x_{start}\) and \(\alpha\)) from the \(n\) gap equations leaves \(n - 2\) independent constraints, one for each consecutive triple of rings:
\[ D_j \equiv E_j \pmod 1, \qquad D_j = y_j - 2y_{j+1} + y_{j+2}, \qquad E_j = \epsilon_j - 2\epsilon_{j+1} + \epsilon_{j+2}, \]
for \(j = 0, \ldots, n-3\). As with the pairwise differences in the vertical case, the vector \((D_0, \ldots, D_{n-3})\) is uniformly distributed on the \((n-2)\)-torus, so the probability of a helix existing is the volume of the region swept out by \((E_0, \ldots, E_{n-3})\) as each \(\epsilon_i\) ranges over \([-f/2, f/2]\) — a zonotope generated by the \(n\) columns of the second-difference operator. Computing that volume (the sum of the absolute values of all \((n-2)\times(n-2)\) minors) gives a binomial coefficient:
\[ P_{\text{helix}}(n) = \binom{n+1}{3} f^{n-2} = \frac{n(n-1)(n+1)}{6} f^{n-2}, \qquad \text{valid for } f \le \tfrac{1}{4}. \]
At \(n = 3\) this is \(\binom{4}{3} f = 4f\), matching the Extra Credit Solution above.
Arbitrary polynomial paths. Both cases above are instances of a single pattern: allow the path down the cylinder to be a degree-\((p-1)\) polynomial helix, \(x(z) \equiv x_0 + a_1 z + \cdots + a_{p-1}z^{p-1} \pmod 1\), which has \(p\) free parameters (\(p=1\) is the vertical line, \(p=2\) is the helix). Eliminating the \(p\) parameters from the \(n\) gap equations leaves \(n-p\) constraints given by the \(p\)-th finite difference of the \(y_i\), and the same torus/zonotope argument gives a closed form for every \(p\) at once:
\[ P(n, p) = \left[ \prod_{1 \le i \le j \le p} \frac{n - p + i + j - 1}{i + j - 1} \right] f^{\,n-p}, \qquad \text{valid for } f \le 2^{-p}. \]
The product collapses to \(n\) when \(p=1\) and to \(\binom{n+1}{3}\) when \(p=2\), recovering \(P_{\text{line}}(n)\) and \(P_{\text{helix}}(n)\) above. Note that the two validity thresholds, \(f \le 1/2\) and \(f \le 1/4\), are likewise the \(p=1\) and \(p=2\) cases of \(f \le 2^{-p}\): more flexible paths raise the success probability but shrink the range of \(f\) over which the closed form holds exactly (beyond that threshold the swept-out region wraps around the torus and starts to self-overlap, so the volume formula overcounts).
Click "Randomize" to rotate the rings. Click and drag the cylinder to rotate the view. The solid red line shows a vertical solution (if one exists), and the dashed green line shows the "best fit" helix (grey if no solution exists). The simulation also runs 10,000 trials to estimate the overall probabilities.